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Math Help - newton's third law + forces

  1. #1
    Senior Member furor celtica's Avatar
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    newton's third law + forces

    three identical boxes, each of weight W, are to be stacked one on top of another against a vertical wall. The lowermost box is in contact with the wall, and the other two boxes are positioned as shown in the diagram. The middle box is pushed into position by the application of a horizontal force of magnitude P.
    Show that if P>4W (where is coefficent of friction between any two boxes), sliding takes place between the upper and middle boxes (i.e. friction is not strong enough for the upper box to follow the middle box in its movement).

    I don't understand why it is 4W, shouldn't it be 3W or 2W? I'm not sure.
    Firstly the middle box must be able to move, i.e. P>(friction from lowermost box), P>(normal reaction) x , P> 2W. However, the frictional force between the upper and middle boxes is only W, os shouldn't the 2W force be enough? Or 3W if both frictional forces are added, I'm not even sure, but any way i can't get to 4W.
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  2. #2
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    You are right: the middle box will start moving when P > 2W. However, if P exceeds 2W just a little, the top box will move along with the middle one; there will be no sliding. Only when P is big enough (note that the problem does not make a upper bound on P), the top box will not be able to keep up with the acceleration of the middle box.

    Assuming the two upper boxes move together, the total force on the middle box is P - 2W, and the acceleration of the upper boxes is (P - 2W) / 2m, where m = W / g. Compare this with the acceleration that the force of friction can exert on the top box.
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  3. #3
    Forum Admin topsquark's Avatar
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    This looks awfully familiar. See here. Please do not do this again.

    See here.


    -Dan
    Last edited by topsquark; February 9th 2011 at 12:07 PM.
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  4. #4
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    Re: newton's third law + forces

    I assume that is the coefficient of static friction.

    The max. horizontal force that the middle book can exert on the top book is W.

    That will give the top book an acceleration of: \displaystyle a = {{\mu W}\over{m_{book}}}={{\mu Wg}\over{W}}}=\mu g

    What force, P, must be exerted on the middle book to make its acceleration greater than μg ?
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